Question

An electron in a hydrogen like atom is in an excited state. it has a total energy of $–3.4eV$. The de Broglie wavelength of the electron is nearly

Answer 1

Step 1: Given data

1. The total energy of the electron is $–3.4eV$

Step 2: de Broglie wavelength

1. The de Broglie wavelength of a particle of mass $m$ and velocity $v$ is defined by the form, $\lambda =\frac{h}{mv}$, where, $h$ is the Plank constant.
2. Another form of de Broglie wave is $\lambda =\frac{h}{\sqrt{2m{E}_k}}$, where, $E_k$ is the kinetic energy of the particle.

Step 3: Calculation of total energy

We know from the concept of energy of the hydrogen atom, the kinetic energy of an electron is negative of total energy.

So, kinetic energy,

$$\begin{gathered} E_k=+3.4eV \\ or{E}_k=3.4\times \left(1.602\times {10}^{-19}\right)Volt. \end{gathered}$$.

Step 4: Calculation of de Broglie wavelength

Now, de Broglie wavelength of the electron is

$$\begin{gathered} \lambda =\frac{h}{\sqrt{2m{E}_k}} \\ \lambda =\frac{6.626\times {10}^{-34}}{\sqrt{\left(2\times 9.1\times {10}^{-31}\times 3.4\times 1.602\times {10}^{-19}\right)}}(plankcons\tan t,h=6.626\times {10}^{-34}) \\ or\lambda =6.653\times {10}^{-10}m. \end{gathered}$$

The de Broglie wavelength of the electron is $6.653\times {10}^{-10}m.$